HW 2 - MATH 290 HOMEWORK 2 SOLUTIONS 1 Prove that for any...

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MATH 290 – HOMEWORK 2 – SOLUTIONS 1. Prove that for any integers a and b , a + b is even if and only if either both a and b are even or both a and b are odd. Solution: Let a and b be integers. (= ) We will prove this direction by contrapositive. Suppose that it is not true that either both a and b are even or both a and b are odd. This means that one of a or b is odd and the other even. Let us assume that a is even and b is odd (the proof is exactly the same if we assume the other way around). Then for some integers k and m , a = 2 k and b = 2 m +1. Then a + b = 2 k +2 m +1 = 2( k + m )+1. Since k + m is an integer, a + b is odd hence not even. ( =) We will prove this by examining two cases. Case 1. Assume that both a and b are even. Then for some integers k and m , a = 2 k and b = 2 m . Then a + b = 2 k +2 m = 2( k + m ). Since k + m is an integer, a + b is even. Case 2. Assume that both a and b are odd. Then for some integers k and m , a = 2 k + 1 and b = 2 m + 1. Then a + b
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HW 2 - MATH 290 HOMEWORK 2 SOLUTIONS 1 Prove that for any...

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